Package of problems

## Problem

For each card below, determine which non-negative values of $a$, $b$, $c$, and $d$, if any, make the equation true. These can be attempted in any order but you might find that some cards can help inform your decisions about others.

$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$

$\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}$

$\sqrt{23-6\sqrt{6-4\sqrt{2}}}=\sqrt{a}+\sqrt{b}$

$a\sqrt{b}=\sqrt{ab}$

$\dfrac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=1$

$\sqrt{a} - \sqrt{b}=\sqrt{a - b}$

$\sqrt{a}+\sqrt{b}=\sqrt{a+b+\sqrt{4ab}}$

$\dfrac{\sqrt{a}+b}{\sqrt{c}+d}=(\sqrt{a}+b)(\sqrt{c}-d)$

$\sqrt{5+2\sqrt{6}}=\sqrt{a}+\sqrt{b}$